Question

The solubility product, K_sp, of Al(OH)₃(s) is 1.0 x 10^-33. What is its solubility (in g/L) in an aqueous solution of NaOH with a pH of 12.30 ?

Answer 1

1.

we have a NaOH with pH=12.30...

Al(OH)3 is dissosiated in water as follows

Al(OH)3------------------>Al⁺³ + OH^-

since this Al(OH)3 is dissolved in NaOH..so due to common ion effect conentration of OH^- of NaOH is same as OH- of Al(OH)3..

let calculate concentration of OH- from NaOH

we have pH = 12.30

we know

pH+pOH =14

pOH=14-pH

pOH =14-12.30

pOH=1.7

and pOH = -log( [OH^-] )

1.7 = -log( [OH^-] )

[OH^-] = 1.995 x 10^-2

now we concentartion of OH- = 1.995 x 10^-2

if the solubility of the compound(Al(OH)₃) is y

the dissociated amount of Al+3 = y

the dissociated amount of 3OH- = 3y

Al(OH)3------------------>Al⁺³ + 3OH^-  

0 1.995 x10^-2 (initial condition)

y (1.995 x 10^-2 + 3y) (final condition)

so Ksp = [ Al⁺³ ] * [ OH^- ] ³

we have Ksp =1.0 x 10^-33

1.0 x 10^-33 = y * (1.995 x 10^-2 + 3y )³

1.0 x 10^-33 = y * (1.995 x 10^-2 )³   (as value of Ksp is very low so value of solubility(3y) would  also be very low)

y = 1.25 x 10^-28 mol/L

as we have to find solubility in gm/L so

moleculer weight of Al(OH)3 = 78 gm/mol

so y = 1.25 x 10^-28 mol/L * 78 gm/mol

y = 97.5 x 10^-28 gm/L